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NEET PHYSICSMAGNETISM AND MATTERMedium

Question

A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in earth's horizontal magnetic field of 24 μT24\ \mu\text{T}. When a horizontal field of 18 μT18\ \mu\text{T} is produced opposite to the earth's field by placing a current-carrying wire, the new time period of the magnet will be:

A

1 s

B

2 s

C

3 s

D

4 s

Step-by-Step Solution

  1. Time Period Formula: The time period (TT) of a magnet oscillating in a uniform magnetic field (BB) is given by T=2πImBT = 2\pi \sqrt{\frac{I}{mB}}, where II is the moment of inertia and mm is the magnetic moment. This implies T1BT \propto \frac{1}{\sqrt{B}}.
  2. Initial Condition:
  • Initial magnetic field (B1B_1) = Earth's horizontal field (HH) = 24 μT24\ \mu\text{T}.
  • Initial time period (T1T_1) = 2 s.
  1. Final Condition:
  • An external field (BextB_{ext}) of 18 μT18\ \mu\text{T} is produced opposite to the Earth's field.
  • Net magnetic field (B2B_2) = HBext=24 μT18 μT=6 μTH - B_{ext} = 24\ \mu\text{T} - 18\ \mu\text{T} = 6\ \mu\text{T}.
  1. Calculation:
  • Using the relation T2T1=B1B2\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}: T22=246=4=2\frac{T_2}{2} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2 T2=2×2=4 sT_2 = 2 \times 2 = 4 \text{ s}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MAGNETISM AND MATTER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMAGNETISM AND MATTERvibrationmagnetometerplacedmagneticmeridian

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